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2 years ago

I need help with this

Chemistry

2 answers:

notka56 [123]2 years ago

8 0

Answer: B

Explanation: All you have to do to find out if an equation is balanced is count the number of atoms on each side. To be balanced, they must be equal.

In case you didn’t know: The number in front of an element or compound means that you will multiply. For instance, 2AlBr3 would have 2 Al atoms and 6 Br atoms. The subscript number just means how many atoms there are in that compound.

marshall27 [118]2 years ago

7 0

What is it you need help on? there is nothing here????

ANSWER:

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2 years ago

When methanol, CH 3 OH , is burned in the presence of oxygen gas, O 2 , a large amount of heat energy is released. For this reas

luda_lava [24]

<u>Answer:</u> The mass of methanol that must be burned is 24.34 grams

<u>Explanation:</u>

We are given:

Amount of heat produced = 581 kJ

For the given chemical equation:

$CH_3OH(g)+\frac{3}{2}O_2(g)\rightarrow CO_2(g)+2H_2O(l);\Delta H=-764kJ$

By Stoichiometry of the reaction:

When 764 kJ of heat is produced, the amount of methanol reacted is 1 mole

So, when 581 kJ of heat will be produced, the amount of methanol reacted will be = $\frac{1}{764}\times 581=0.7605mol$

To calculate mass for given number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Moles of methanol = 0.7605 moles

Molar mass of methanol = 32 g/mol

Putting values in above equation, we get:

$0.7605mol=\frac{\text{Mass of methanol}}{32g/mol}\\\\\text{Mass of methanol}=(0.7605mol\times 32g/mol)=24.34g$

Hence, the mass of methanol that must be burned is 24.34 grams

4 0

3 years ago

What is the final volume of a gas (in liters) when the initial volume is 14.00 L at

8_murik_8 [283]

Answer:

$V_2=1344L$

Explanation:

Hello there!

In this case, since we have a problem about volume-pressure relationship, the idea here is to use the Boyle's law to calculate the final volume as shown below:

$P_2V_2=P_1V_1\\\\V_2=\frac{P_2V_2}{P_1}\\$

Then, we plug in the initial and final pressures and the initial volume to obtain:

$V_2=\frac{14.00L*0.9600atm}{0.01000atm}\\\\V_2=1344L$

Regards!

4 0

2 years ago